CHEMISTRY COMMUNITY

Yuelai Feng 3E

Hi! Zinc ion is usually 2+ charged because its electron configuration is [Ar]3d 10 4s 2 , so it readily loses 2 valence electrons in 4s-orbital, similar to Group 2 metals. Then, since hydroxide ion and nitrite ion are both 1- charged, the "zinc:anion" ratio is 1:2. This gives you Zn(OH) 2 ...

Yuelai Feng 3E

In my understanding, K a is only dependent on the type of acid, while pH is dependent on both the type of substance and the concentration of the solution as well. K a tells you the extent to which an acid dissociates into H+ and A-. ( K_a = \frac{[H^+][A^-]}{[HA]} ) Strong acids almost completely di...

Yuelai Feng 3E

Hi! Suppose the initial molar concentration of HCl is x (mol/L), then the final concentration is 0.12x (mol/L). Because HCl is a strong acid, we assume that it completely dissociates into H+ and Cl- in the solution. Therefore: - In the initial solution, [H+] initial =[Cl-] initial =x (mol/L); - In t...

Yuelai Feng 3E

Hi! In the Bronsted-Lowry definition, an acid is a proton donor and a base is a proton acceptor. In the Lewis definition, an acid is a electron acceptor and a base is an electron donor. (The Lewis definition has a wider range of application than Bronsted, but in many cases the Bronsted definition is...

Yuelai Feng 3E

Hi! The question asks us to draw ICl₃ Here's a screenshot of my textbook.

Yuelai Feng 3E

Hi! Here are the general guidelines for finding oxidation states: https://www.chemguide.co.uk/inorganic/r ... tates.html
The oxidation state of a ligand is equal to its charge; and the sum of oxidation states in a coordination compound is equal to its overall charge. Hope it helps!

Yuelai Feng 3E

Hi! Like what the previous answer says, sigma bonds can rotate while the existence of pi bonds restrict rotation. So a ligand (with several lone pair sites) that consists of all single bonds can probably "get around" and form multiple bonds at the same time. Also, there should be enough sp...

Yuelai Feng 3E

Hi! In the lecture Dr. Lavelle gave an example [Co(NH3)5Cl], where NH3 is placed before Cl. So I guess the order of the ligands doesn't really matter, as long as the TM cation is the first one it should be fine!

Yuelai Feng 3E

Hi! For part a, HN(CH2CH2NH2)2 is tridentate not only because it has 3 N atoms with a lone pair that can form a coordinate bond, but also because it consists of all single bonds, which can rotate and keep the 3 lone pairs next to the central TM cation. For part b, all 3 oxygen atoms in CO 3 2- has l...

Yuelai Feng 3E

Hi! The structure shown on the textbook didn't draw the lone pair electrons. For part b, the oxygen has 2 lone pairs besides the 2 bonds shown in the graph, so there are 4 electron density regions in total. Therefore, the shape of electron regions is tetrahedral, and the molecular shape is bent. The...

Yuelai Feng 3E

For the lewis structure of I 3 - , the answer says there are 2 single bonds joining the iodine atoms, and the central atom has 3 lone pairs. But what if there is 1 single bond & 1 double bond, and the central atom has 2 lone pairs? I'm not sure why this is less valid than the correct answer, sin...

Yuelai Feng 3E

For the central C atom in [H 2 -C=C=C-H 2 ] (call it C2), it has two unhybridized p-orbitals, forming two pi bonds. Therefore, in order for the two p orbitals to co-exist and minimize electron repulsion, they must be perpendicular to each other. Then, for the two neighboring C atoms (call them C1 an...

Yuelai Feng 3E

Hi! I believe all molecules with bent structure should be polar. Even if all three atoms are the same -- for example, O3 -- the molecule is still polar. The lone pair electrons and bonding electrons are arranged asymmetrically around the central atom, so the dipole moments do not cancel out.

Yuelai Feng 3E

Hi! Because for the Carbon atom in the middle, it has 2 pi bonds formed by two 2p-orbitals. These two orbitals must be perpendicular to one another to have minimum electron repulsion (just like how px, py, and pz-orbitals are arranged in an atom). Hope it helps.

Yuelai Feng 3E

https://chem.libretexts.org/Courses/City_College_of_San_Francisco/Chemistry_101A/Topic_E%3A_Atomic_Structure/07%3A_Electronic_Structure_of_Atoms/7.05%3A_Quantum_Mechanics_and_Atomic_Orbitals I think this is a pretty straightforward explanation of how wave functions define atomic orbitals. Hope it he...

Yuelai Feng 3E

1. Yes, the full orbital means that the lone pair will not form a bond. It will remain as a lone pair, as you see in the Lewis structure of NH3. 2. I think the hybridization model can be applied to all molecules. It's used to explain their observed structure by making several e- orbitals have equal ...

Yuelai Feng 3E

Charlotte Adams 2D wrote:So if there are more electrons, is the induced-dipole induced-dipole stronger and harder to break?

Thanks

Yes you're right. But induced-dipole induced-dipole interactions are all very weak compared to other interactions.

Yuelai Feng 3E

Hi! Elements in period 3 or higher can have an expanded octet. When you draw lewis structures, if the central atom is has period >= 3 and the surrounding atoms have not yet fulfill their octet, you should consider expanding the octet of the central atom and turn some single bonds into double bonds. ...

Yuelai Feng 3E

Hi! Cations with higher polarizing power (which are smaller & with higher charge density) tend to form bonds with more covalent characters. This is because the cation exerts higher electrostatic attraction on the electrons surrounding the anion, causing them to delocalize and form something simi...

Yuelai Feng 3E

Hi! In the melted form (and even gaseous form), there is still attraction force between Na+ and Cl- ions, but the kinetic energy is too high to let them maintain a lattice structure like solid NaCl does. I think you are mostly right. On the microscopic level there can be Na+Cl- momentarily closely a...

Yuelai Feng 3E

Hi! This is because having a full 3d sub-shell makes the atom more stable with lower energy. Similarly, Cr has [Ar] 3d5 4s1 instead of [Ar] 3d4 4s2, since having half-full 3d also lowers the energy. Hope it helps!

Yuelai Feng 3E

Hi! Not sure if I understand your question correctly. This might differ from case to case, but if the surrounding atoms are different, they will usually have different electronegativity, and the structures in which "more electronegative atoms having more negative formal charges" will be mo...

Yuelai Feng 3E

That was a really nice summary!! Just to reorganize this a bit, I think the 2 major factors are Polarizability and Distance, and all others are secondary factors that influence the major factors. Factors affecting induced dipole-induced dipole interactions Ep∝(ɑ1 ɑ2)/(r^6) 1. Polarizability (∝ Elect...

Yuelai Feng 3E

Hi! I think it's easier to explain this with a graph.

Wherever there's an H attached to N/O/F, it can form H-bond with the lone pair on O atom in H2O.
Whenever there's a lone pair on an electronegative atom (N/O/F), it can form a H-bond with the H in H2O.
Hope this helps!

Yuelai Feng 3E

Hi! Theoretically speaking the max. # of valence electrons is the max. "capacity" of the outermost electron shell. For example, Period 3 elements can have a maximum of 18 valence electrons. Because the third electron shell can hold 18 e- (2s+6p+10d). For Period 4 elements it should be 2s+6...

Yuelai Feng 3E

Hi! So in the lewis structure you suggested, the sum of |formal charge| is 2, which is the same as the 3 structures in the solution manual. In this case, we need to compare the electronegativity of the atoms to determine which structure is more valid. Here, O is more electronegative than P, so O wil...

Yuelai Feng 3E

Hi! I did this question and the answer should be 8. One hydrogen bond can be formed for each N-H bond, one for each N atom (since each has 1 lone pair e-), and two for the O atom (since it has 2 lone pairs of e-). According to wikipedia, a hydrogen bond is "a primarily electrostatic force of at...

Yuelai Feng 3E

Hi! You only need the first 3 quantum numbers to specify an orbital (in which there're 2 electrons), and you'll need to include m_s to specify which electron it is.

Yuelai Feng 3E

Hi! Basically in the periodic table, every block that precedes a specific atom stands for an electron it has, and it is listed in a sequence as if the electron orbitals are "filled up" one by one: e.g. 4s is filled before 3d. However, valence electrons only include electrons with the large...

Yuelai Feng 3E

Hi! For b and c, you can compare the electronegativity of each atom. In this case, O is more electronegative than N, and N is more electronegative than C. Therefore, O taking electrons from N (which is b) is more likely to happen than C taking electrons from N (which is c). Negative formal charges u...

Yuelai Feng 3E

Hi! I guess you're thinking that the electrons in d-orbital are the valence electrons. But actually the valence e- includes all electrons in the 3rd shell (n=3). So the maximum # of electrons is s2 + p6 + d10 = 18 e-. (But it doesn't mean they have to get 18 electrons. For period 3 elements it's usu...

Yuelai Feng 3E

Yes, just like Isabella said, elements in period 3 or higher can have expanded octet, and it is their stable form. Dr. Lavelle mentioned in the lecture that in these cases, the expanded octet has lower energy than octet structures, so they are more stable.

Yuelai Feng 3E

Hi! The maximum number of valence electrons depends on the electron orbitals the element can have. For elements with n>=3, they can have more than 8 electrons because they can have s2 + p6 + d10 = 18 electrons. For elements with n>=4, theoretically they can have s2 + p6 + d10 + f14 = 32 electrons ma...

Yuelai Feng 3E

Hi! If the energy of light matches the energy difference between two electron shells, then the energy will be absorbed and it will excite the electron to a higher energy level. This is the same situation as in atomic absorption spectrum. However, if the energy does not match then it will not be abso...

Yuelai Feng 3E

Hi! I usually put the 2 units I want to convert into a fraction. For example, I know that 1m=100cm, so the fraction can be \frac{1m}{100cm} or \frac{100cm}{1m} . Because the denominator and numerator are the same, both fractions equal to 1. Then, I can multiply a value by this fraction, and cancel o...

Yuelai Feng 3E

Hi! In terms of solving homework problems I have only used '3.28984 x10^15 Hz' so far. Perhaps the other value is more useful when solving other types of questions which we have not yet encounter. :)

Yuelai Feng 3E

Hi! I don't think they can all be the same because the acceptable numerical values for each is different. For example, 'n' should be a natural number like 1, 2, 3, 4 ...; 'l' should be a non-negative integer from zero to (n-1); and 'ml' should be an integer from -l to l. So n and l can never be the ...

Yuelai Feng 3E

Hi! I think the 3 quantum numbers are three parameters of wave function, which is the solution of Schrodinger equation. The Schrodinger Equation can determine the wave function (i.e. orbital) of an electron in an atom, and the wave function specifies the behavior of the electron wave in relation to ...

Yuelai Feng 3E

Hi! I think the main problem here is that E=hc/λ gives you the total energy of the wave, not just the kinetic energy. So you need to calculate the momentum (and thus velocity) of the electron first, and then work out Ek with 1/2mv^2. Also, the De Broglie's equation only works for particle with mass,...

Yuelai Feng 3E

It's me again. I guess I kind of figured it out a little bit... Is it that we can derive the wave functions if we impose boundary conditions to them (for example, in 1 dimension, we can limit it in a length L), therefore only certain wavelengths are acceptable, and then we can determine all possible...

Yuelai Feng 3E

Hi! I know that Schrodinger Equation can be used to calculate the energy of electrons given their wave functions. But I am unsure about how the wave functions (like the s, p, d, f orbitals) were derived in the first place -- I mean, in order to determine those wave functions, we need to know the emp...

Yuelai Feng 3E

I don't have a confident answer to this, but it might be helpful to think about "what is waving" there. For sound waves, it's the particles of the media; for EM waves it's electromagnetic fields; and for particles like electrons, it might be something else. Hope it helps!

Yuelai Feng 3E

Hi! You can think of the Hamiltonian operator as a kind of "template" or "equation". You plug in the wave function psi and get the result (E psi) This is a screenshot of the textbook: 截屏2020-10-22 下午2.29.10.png The left hand side of the equation (H-psi) is what we get after putti...

Yuelai Feng 3E

Hi! I was wondering about these questions too. What I learned from the textbook and YouTube videos is that the wave function (psi) itself doesn't really have a physical meaning, at least we haven't figured out how to interpret it. So I guess it's just a mathematical representation of "an electr...

Yuelai Feng 3E

Hi! I believe frequency is continuous for both wave and particle model. (like when we use v=E/h to calculate frequency of photons, and the result can be a continuous range of values) However, in the wave model energy transfer is continuous (i.e. particles can absorb any amount of energy from light);...

Yuelai Feng 3E

Hi! For Q29, the work function for sodium is 150.6 kJ per mol, so you need to divide 150.6 kJ/mol by Avogadro's constant (6.02*10^23) to get the answer (it should be 2.501x10^-19 J). And for Q30, you should also replace the threshold energy with the answer you get from Q28, since E=hv represents the...

Yuelai Feng 3E

Hi! I think the UV region refers to the light with wavelength 10-400 nm. The image below is a spectrum showing the different regions of light. They are categorized and named according to their wavelength/frequency. Hope it helps!

Yuelai Feng 3E

Hi! In the photon model, light intensity is determined by the number of photons; and for each photon, its energy is determined by its frequency (E=hv). In the wave model, light intensity is determined by the amplitude of the wave, and energy is correlated with both amplitude and frequency. In other ...

Yuelai Feng 3E

I think whether you add the negative sign or not depends on the values of n1 and n2. You just have to make sure the frequency (v) is positive. My understanding is: The Rydberg equation is essentially derived from E_n=-\frac{hR}{n^2} . The energy difference between two energy levels can be expressed ...

Yuelai Feng 3E

Hi! So after converting the wavelength into frequency using c=λv, you can plug in the frequency (v) to the Rydberg equation to determine n2. 0.112 is the value of \frac{1}{(n_2)^2} , so (n_2)^2 is equal to \frac{1}{0.112} (approximately =9). Therefore (n_2) is \sqrt\frac{1}{0...

Yuelai Feng 3E

Hi! I'm currently enrolled in CHEM 14A. While planning my future courses, I'm wondering if CHEM 14B and MATH 31B have to be taken together? (It shows MATH 31B is the co-requisite, not pre-requisite) Is it possible to take MATH 31B first, and CHEM 14B later on?
Thank you!

Yuelai Feng 3E

Hi! I'm not entirely sure of this, but I think the equation c=λv can be derived from the units. Wavelength (λ) is the distance between two peaks (measured in meter), and frequency (v) is the number of reps in each second (measured in second^-1). So if you multiply the two units you'll get m*s^-1, wh...

Yuelai Feng 3E

Hi! The reason why NaOH should be the limiting reagent, not Cu(NO3)2, is that they react in a 2:1 ratio. The chemical equation is: Cu(NO3)2 + 2 NaOH ---> Cu(OH)2 + 2NaNO3 Therefore, for 0.04 mol Cu(NO3)2, 0.08 mol NaOH is needed, but we only have 0.05 mol NaOH. So actually NaOH is the limiting reage...

Yuelai Feng 3E

Hi! For this question, we can first calculate the masses of the product and reactants. Given the crucible is 26.45 g, and the crucible + product are 28.35 g in total, the mass of the product (the oxide) is 28.35g-26.45g = 1.90g. And we also know that the tin sample is 1.50g. So we have: Sn (1.50g) +...

Yuelai Feng 3E

Hi! I've posted a response under another question that asked this problem, so I'll just paste it here: --- First, we can calculate the number of molecules in the 10mL original solution: n_{initial} = c_{initial} \cdot V = 0.1 \, mol/L \cdot 10\,mL = 0.0010\,mol Then, each time the volume is doubled,...

Yuelai Feng 3E

Hi! we only care about molar ration when dealing with empirical formulas. The mass percent composition can be seen as the mass ratio of different elements in a sample. The ratio won't be affected by sample mass. And I want to add that when we want to find the molecular formula when given an empiric...

Yuelai Feng 3E

Hi! That's because, after you convert the mass of each element into the number of moles, you still need to divide each mole number by the smallest one to get the empirical formula. Only the molar ratio matters. For example, if you imagine there are 100 grams of sample, and you get 1mol C + 4mol H. T...

Yuelai Feng 3E

A stimulant in coffee and tea is caffeine, a substance of molar mass 194 g/mol -1. When 0.376 g of caffeine was burned, 0.682 g of carbon dioxide, 0.174 g of water, and 0.110 g of nitrogen were formed. Determine the empirical and molecular formulas of caffeine. The unbalanced equation is: Caffeine ...

Yuelai Feng 3E

I don't think we need to know all the conventional names for complex molecules like fluoxetine. (But for very common ones, like glucose, it might be helpful to know.) This is just a simple way to refer to the molecule but has little to do with its molecular structure. The IUPAC name is the one that ...

Yuelai Feng 3E

First, we can calculate the number of molecules in the 10mL original solution: n_{initial} = c_{initial} \cdot V = 0.1 \, mol/L \cdot 10\,mL = 0.0010\,mol Then, each time the volume is doubled, the concentration of the solution becomes one half, and the number of molecules in 10mL of the resultant s...

Yuelai Feng 3E

Hi! The equation for that question should be 2A+1B—> 3 C, not 2A+1B—>C. The "maximum amount of product" is another way of saying the "theoretical yield". In this case, you have 1 mol A reacting with 0.5 mol B, leaving the other 0.5 mol B in excess. The question is asking for the ...

Search found 61 matches